A polar covalent bond has an asymmetric electron distribution due to a difference in electronegativity.
Inductive effect:
Shift of electron distribution in a σ σ sigma\sigmaσ bond in response to a difference in electronegativity.
Dipole moment : μ ― = Q r ― μ _ = Q r _ mu _=Qr_\underline{\mu}=\mathrm{Q} \underline{r}μ―=Qr―
Na + Cl − Na + Cl − Na^(+)Cl^(-)\mathrm{Na}^{+} \mathrm{Cl}^{-}Na+Cl− 9.0 D
CH 3 Cl CH 3 Cl CH_(3)Cl\mathrm{CH}_{3} \mathrm{Cl}CH3Cl 1.87 D
H 2 O H 2 O H_(2)O\mathrm{H}_{2} \mathrm{O}H2O 1.85 D
NH 3 NH 3 NH_(3)\mathrm{NH}_{3}NH3 1.47 D
CH 4 CH 4 CH_(4)\mathrm{CH}_{4}CH4 0
CO 2 CO 2 CO_(2)\mathrm{CO}_{2}CO2 0
Dipole moment : mu _=Qr_ Na^(+)Cl^(-) 9.0 D CH_(3)Cl 1.87 D H_(2)O 1.85 D NH_(3) 1.47 D CH_(4) 0 CO_(2) 0| Dipole moment : | $\underline{\mu}=\mathrm{Q} \underline{r}$ | | :---: | :---: | | | | | $\mathrm{Na}^{+} \mathrm{Cl}^{-}$ | 9.0 D | | $\mathrm{CH}_{3} \mathrm{Cl}$ | 1.87 D | | $\mathrm{H}_{2} \mathrm{O}$ | 1.85 D | | $\mathrm{NH}_{3}$ | 1.47 D | | $\mathrm{CH}_{4}$ | 0 | | $\mathrm{CO}_{2}$ | 0 |
bond dipoles
O = C = O + m ― = 0 cancellation O = C = O + m _ = 0  cancellation  O=C=O^(+)quad{:[m_=0],[" cancellation "]:}\mathrm{O}=\stackrel{+}{\mathrm{C}=\mathrm{O}} \quad \begin{gathered} \underline{\mathrm{m}}=0 \\ \text { cancellation } \end{gathered}O=C=O+m―=0 cancellation 
CC(C)=OO
The formal charge on an atom is a method for electron bookkeeping. Chemical reactions involve changes in bonding, and it is useful to have a convention for counting electrons to help understand and predict chemical reactions. Our counting convention consists of formal charge plus arrowpushing (discussed later in this chapter).
Formal charge = ( No. of valence electrons in the free atom ) − ( No. of valence electrons in the bound atom ) = ( No. of valence electrons in the free atom ) − 1 2 ( No. of shared electrons ) − ( No. of nonbonding (lone pair) electrons ) NH 3 Formal charges of NH 3 N has 5 valence electrons H has 1 valence electron N 5 − 1 2 ( 6 ) − 2 = 0 H 1 − 1 2 ( 2 ) − 0 = 0  Formal charge  = (  No. of valence electrons   in the free atom  ) − (  No. of valence electrons   in the bound atom  ) = (  No. of valence electrons   in the free atom  ) − 1 2 (  No. of shared   electrons  ) − (  No. of nonbonding   (lone pair) electrons  ) NH 3  Formal charges of  NH 3 N  has  5  valence electrons  H  has  1  valence electron  N 5 − 1 2 ( 6 ) − 2 = 0  H  1 − 1 2 ( 2 ) − 0 = 0 {:[" Formal charge "=((" No. of valence electrons ")/(" in the free atom "))-((" No. of valence electrons ")/(" in the bound atom "))],[=((" No. of valence electrons ")/(" in the free atom "))-(1)/(2)((" No. of shared ")/(" electrons "))-((" No. of nonbonding ")/(" (lone pair) electrons "))],[{:[NH_(3)," Formal charges of "NH_(3)],[N" has "5" valence electrons "],[H" has "1" valence electron "]:}Nquad5-(1)/(2)(6)-2=0],[" H "quad1-(1)/(2)(2)-0=0]:}\begin{aligned} \text { Formal charge } & =\binom{\text { No. of valence electrons }}{\text { in the free atom }}-\binom{\text { No. of valence electrons }}{\text { in the bound atom }} \\ & =\binom{\text { No. of valence electrons }}{\text { in the free atom }}-\frac{1}{2}\binom{\text { No. of shared }}{\text { electrons }}-\binom{\text { No. of nonbonding }}{\text { (lone pair) electrons }} \\ \begin{array}{cl} \mathrm{NH}_{3} & \text { Formal charges of } \mathrm{NH}_{3} \\ \mathrm{~N} \text { has } 5 \text { valence electrons } \\ \mathrm{H} \text { has } 1 \text { valence electron } \end{array} & \mathbf{N} \quad 5-\frac{1}{2}(6)-2=0 \\ & \text { H } \quad 1-\frac{1}{2}(2)-0=0 \end{aligned} Formal charge =( No. of valence electrons  in the free atom )−( No. of valence electrons  in the bound atom )=( No. of valence electrons  in the free atom )−12( No. of shared  electrons )−( No. of nonbonding  (lone pair) electrons )NH3 Formal charges of NH3 N has 5 valence electrons H has 1 valence electron N5−12(6)−2=0 H 1−12(2)−0=0
C=C(C)C(C)C
Molecular charge = 0 = 0 =0=0=0
C ( CH 3 ) 4 − 1 2 ( 8 ) − 0 = 0 C CH 3 4 − 1 2 ( 8 ) − 0 = 0 C(CH_(3))quad4-(1)/(2)(8)-0=0\mathrm{C}\left(\mathrm{CH}_{3}\right) \quad 4-\frac{1}{2}(8)-0=0C(CH3)4−12(8)−0=0
O 6 − 1 2 ( 6 ) − 2 = + 1 6 − 1 2 ( 6 ) − 2 = + 1 6-(1)/(2)(6)-2=+16-\frac{1}{2}(6)-2=+16−12(6)−2=+1
C: 4 − 1 2 ( 6 ) − 2 = − 1 4 − 1 2 ( 6 ) − 2 = − 1 quad4-(1)/(2)(6)-2=-1\quad 4-\frac{1}{2}(6)-2=-14−12(6)−2=−1
How to draw it:
CC(C)=[Ge](C)CGe

or
CC(C)[O+](C)CO+

C[Si](C)C(C)(C)[Si](C)(C)CSiSi
Nitromethane (Molecular charge = 0)
CN([O])[O+]NOO+
N 5 − 1 2 ( 8 ) − 0 = + 1 5 − 1 2 ( 8 ) − 0 = + 1 quad5-(1)/(2)(8)-0=+1\quad 5-\frac{1}{2}(8)-0=+15−12(8)−0=+1
O 1 O 1 O_(1)\mathbf{O}_{1}O1
6 − 1 2 ( 4 ) − 4 = 0 6 − 1 2 ( 4 ) − 4 = 0 6-(1)/(2)(4)-4=06-\frac{1}{2}(4)-4=06−12(4)−4=0
O2 6 − 1 2 ( 2 ) − 6 = − 1 6 − 1 2 ( 2 ) − 6 = − 1 quad6-(1)/(2)(2)-6=-1\quad 6-\frac{1}{2}(2)-6=-16−12(2)−6=−1
How to draw it:
C[N+](=O)O[Te]N+OOTe

or
C[N+](=O)[O-]N+OO-
Line-bond structures with Lewis lone pairs are adequate for localized bonding, but not for molecules like nitromethane that have delocalized pi bonding. The unsymmetrical charge distribution above cannot be correct, because nitromethane is known to be symmetric with equal N-O bond lengths. For a proper description of the nitromethane wavefuction using line-bond structures and lone pairs, we need an equally weighted resonance hybrid of the two unsymmetrical structures:
The symbol " ⟷ ⟷ longleftrightarrow\longleftrightarrow⟷ " means resonance, not equilibrium.
The wavefunction of nitromethane is: Ψ nitromethane = 1 2 ( Φ 1 + Φ 2 ) Ψ nitromethane  = 1 2 Φ 1 + Φ 2 Psi_("nitromethane ")=(1)/(sqrt2)(Phi_(1)+Phi_(2))\Psi_{\text {nitromethane }}=\frac{1}{\sqrt{2}}\left(\Phi_{1}+\Phi_{2}\right)Ψnitromethane =12(Φ1+Φ2). Resonance structures Φ 1 Φ 1 Phi_(1)\Phi_{1}Φ1 and Φ 2 Φ 2 Phi_(2)\Phi_{2}Φ2 are not independent molecules. They are different bonding descriptions of one molecule, and resonance theory gives the electron distribution in that molecule. In the resonance hybrid Ψ nitromethane Ψ nitromethane  Psi_("nitromethane ")\Psi_{\text {nitromethane }}Ψnitromethane , the O atoms share the negative charge equally, the N-O bonds are intermediate between single and double, and the N-O bond lengths are the same. This gives the correct symmetry of the molecule.
Because resonance structures refer a single molecule, the nuclear positions must be identical for all resonance structures.
Conjugation refers to a delocalized π π pi\piπ system with parallel p orbitals on adjacent carbons. A conjugated 3-atom π π pi\piπ system (an allylic group) is shown below. The X , Y , Z X , Y , Z X,Y,Z\mathrm{X}, \mathrm{Y}, \mathrm{Z}X,Y,Z atoms are sp 2 sp 2 sp^(2)\mathrm{sp}^{2}sp2 hybridized.

Using curved arrows to generate resonance structures

Move electron pairs, not atoms
Start the arrow at the position with the strongest electron-donating ability
  1. At a lone pair on an atom with a negative charge
  2. At a lone pair on a neutral atom
  3. At a pi bond


In CH 3 NO 2 CH 3 NO 2 CH_(3)NO_(2)\mathrm{CH}_{3} \mathrm{NO}_{2}CH3NO2 and the allyl group, the resonance structures have equal weight in the resonance hybrid. In the general case, resonance structures have unequal weights.
How to identify the major resonance structure, in order of importance:
  1. Octet rule
  2. Minimum charge separation
  3. Negative charge should be on the more electronegative atom
Brønsted Acid-Base Equilibria
Acid strength is measured by pK a pK a pK_(a)\mathrm{pK}_{\mathrm{a}}pKa
CH 3 CO 2 H + H 2 O ⇌ CH 3 CO 2 − + H 3 O + K eq = [ CH 3 CO 2 − ] [ H 3 O + ] [ CH 3 CO 2 H ] [ H 2 O ] [ H 2 O ] K eq = [ CH 3 CO 2 − ] [ H 3 O + ] [ CH 3 CO 2 H ] = K a ( CH 3 CO 2 H ) = 0.000025 pK a ( CH 3 CO 2 H ) = − log ⁡ K a ( CH 3 CO 2 H ) = 4.6 CH 3 CO 2 H + H 2 O ⇌ CH 3 CO 2 − + H 3 O + K eq = CH 3 CO 2 − H 3 O + CH 3 CO 2 H H 2 O H 2 O K eq = CH 3 CO 2 − H 3 O + CH 3 CO 2 H = K a CH 3 CO 2 H = 0.000025 pK a CH 3 CO 2 H = − log ⁡ K a CH 3 CO 2 H = 4.6 {:[CH_(3)CO_(2)H+H_(2)O⇌CH_(3)CO_(2)^(-)+H_(3)O^(+)],[K_(eq)=([CH_(3)CO_(2)^(-)][H_(3)O^(+)])/([CH_(3)CO_(2)H][H_(2)O])quad[H_(2)O]K_(eq)=([CH_(3)CO_(2)^(-)][H_(3)O^(+)])/([CH_(3)CO_(2)H])=K_(a)(CH_(3)CO_(2)H)],[=0.000025],[pK_(a)(CH_(3)CO_(2)H)=-log K_(a)(CH_(3)CO_(2)H)],[=4.6]:}\begin{aligned} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} & \\ \mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]} \quad\left[\mathrm{H}_{2} \mathrm{O}\right] \mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]} & =\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right) \\ & =0.000025 \\ \mathrm{pK}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right) & =-\log \mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right) \\ & =4.6 \end{aligned}CH3CO2H+H2O⇌CH3CO2−+H3O+Keq=[CH3CO2−][H3O+][CH3CO2H][H2O][H2O]Keq=[CH3CO2−][H3O+][CH3CO2H]=Ka(CH3CO2H)=0.000025pKa(CH3CO2H)=−log⁡Ka(CH3CO2H)=4.6
In an arbitrary solvent, K a ( CH 3 CO 2 H ) K a CH 3 CO 2 H K_(a)(CH_(3)CO_(2)H)\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)Ka(CH3CO2H) is the effective equilibrium constant for the following:
CH 3 CO 2 H ⇌ CH 3 CO 2 − + H + K a ( CH 3 CO 2 H ) = [ CH 3 CO 2 − ] [ H + ] [ CH 3 CO 2 H ] CH 3 CO 2 H ⇌ CH 3 CO 2 − + H + K a CH 3 CO 2 H = CH 3 CO 2 − H + CH 3 CO 2 H CH_(3)CO_(2)H⇌CH_(3)CO_(2)^(-)+H^(+)quadK_(a)(CH_(3)CO_(2)H)=([CH_(3)CO_(2)^(-)][H^(+)])/([CH_(3)CO_(2)H])\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}+\mathrm{H}^{+} \quad \mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)=\frac{\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]}CH3CO2H⇌CH3CO2−+H+Ka(CH3CO2H)=[CH3CO2−][H+][CH3CO2H]
H + H + H^(+)\mathrm{H}^{+}H+means protonated solvent ( = H 3 O + = H 3 O + =H_(3)O^(+)=\mathrm{H}_{3} \mathrm{O}^{+}=H3O+in water).
Which side of an acid-base equilibrium is present in higher concentration? (In other words, which side has the lower free energy?)
Procedure: compare pK a pK a pK_(a)\mathrm{pK}_{\mathrm{a}}pKa values of the conjugate acids.
The molecules with the highest concentration are the weakest acid and weakest base (in other words, the most thermodynamically stable species).
Derivation of K eq = 10 DpKa K eq  = 10 DpKa K_("eq ")=10DpKa\mathrm{K}_{\text {eq }}=10 \mathrm{DpKa}Keq =10DpKa
Consider K a K a K_(a)\mathrm{K}_{\mathrm{a}}Ka for the two acids:
CH 3 NH 3 + ⇌ CH 3 NH 2 + H + Ka ( CH 3 NH 3 + ) = [ CH 3 NH 2 ] [ H + ] [ CH 3 NH 3 + ] CH 3 CO 2 H ⇌ CH 3 CO 2 − + H + K a ( CH 3 CO 2 H ) = [ CH 3 CO 2 − ] [ H + ] [ CH 3 CO 2 H ] CH 3 NH 3 + ⇌ CH 3 NH 2 + H + Ka CH 3 NH 3 + = CH 3 NH 2 H + CH 3 NH 3 + CH 3 CO 2 H ⇌ CH 3 CO 2 − + H + K a CH 3 CO 2 H = CH 3 CO 2 − H + CH 3 CO 2 H {:[CH_(3)NH_(3)^(+)⇌CH_(3)NH_(2)+H^(+),Ka(CH_(3)NH_(3)^(+))=([CH_(3)NH_(2)][H^(+)])/([CH_(3)NH_(3)^(+)])],[CH_(3)CO_(2)H⇌CH_(3)CO_(2)^(-)+H^(+),K_(a)(CH_(3)CO_(2)H)=([CH_(3)CO_(2)^(-)][H^(+)])/([CH_(3)CO_(2)H])]:}\begin{array}{ll} \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{H}^{+} & \mathrm{Ka}\left(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right)=\frac{\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]} \\ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}+\mathrm{H}^{+} & \mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)=\frac{\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]} \end{array}CH3NH3+⇌CH3NH2+H+Ka(CH3NH3+)=[CH3NH2][H+][CH3NH3+]CH3CO2H⇌CH3CO2−+H+Ka(CH3CO2H)=[CH3CO2−][H+][CH3CO2H]
K eq = [ CH 3 NH 2 ] [ CH 3 CO 2 H ] [ CH 3 NH 3 + ] [ CH 3 CO 2 − ] (by definition) = [ CH 3 NH 2 ] [ H + ] [ CH 3 CO 2 H ] [ CH 3 NH 3 + ] [ CH 3 CO 2 − ] [ H + ] = K a ( CH 3 NH 3 + ) K a ( CH 3 CO 2 H ) = 10 − pK a ( CH 3 NH 3 + ) 10 − pK a ( CH 3 CO 2 H ) = 10 − ( 11 − 5 ) = 10 − 6 ( pK a = − log ⁡ K a ) K eq = CH 3 NH 2 CH 3 CO 2 H CH 3 NH 3 + CH 3 CO 2 − (by definition)  = CH 3 NH 2 H + CH 3 CO 2 H CH 3 NH 3 + CH 3 CO 2 − H + = K a CH 3 NH 3 + K a CH 3 CO 2 H = 10 − pK a CH 3 NH 3 + 10 − pK a CH 3 CO 2 H = 10 − ( 11 − 5 ) = 10 − 6 pK a = − log ⁡ K a {:[K_(eq)=([CH_(3)NH_(2)][CH_(3)CO_(2)H])/([CH_(3)NH_(3)^(+)][CH_(3)CO_(2)^(-)])(by definition) ],[=([CH_(3)NH_(2)][H^(+)][CH_(3)CO_(2)H])/([CH_(3)NH_(3)^(+)][CH_(3)CO_(2)^(-)][H^(+)])=(K_(a)(CH_(3)NH_(3)^(+)))/(K_(a)(CH_(3)CO_(2)H))=(10^(-pK_(a)(CH_(3)NH_(3)^(+))))/(10^(-pK_(a)(CH_(3)CO_(2)H)))=10^(-(11-5))=10^(-6)],[(pK_(a)=-log K_(a))]:}\begin{aligned} \mathrm{K}_{\mathrm{eq}}= & \frac{\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]}{\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]} \text {(by definition) } \\ = & \frac{\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]}{\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right]\left[\mathrm{H}^{+}\right]}=\frac{\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right)}{\mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)}=\frac{10^{-\mathrm{pK}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right)}}{10^{-\mathrm{pK}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)}}=10^{-(11-5)}=10^{-6} \\ & \left(\mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right) \end{aligned}Keq=[CH3NH2][CH3CO2H][CH3NH3+][CH3CO2−](by definition) =[CH3NH2][H+][CH3CO2H][CH3NH3+][CH3CO2−][H+]=Ka(CH3NH3+)Ka(CH3CO2H)=10−pKa(CH3NH3+)10−pKa(CH3CO2H)=10−(11−5)=10−6(pKa=−log⁡Ka)
H 2 O + HCl ⇌ H 3 O + + Cl − H 2 O + HCl ⇌ H 3 O + + Cl − H_(2)O+HCl⇌H_(3)O^(+)+Cl^(-)\mathrm{H}_{2} \mathrm{O}+\mathrm{HCl} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}H2O+HCl⇌H3O++Cl−
Use curved arrows to depict the changes in bonding and keep track of electron pairs in reactions
Move electron pairs, not atoms
Start the arrow at the position with the strongest electron-donating ability
  1. At a lone pair on an atom with a negative charge
  2. At a lone pair on a neutral atom
  3. At a pi bond
Forward reaction
Backward reaction
Wrong
Cl[Hg]C[C@@H]1[CH]O1ClHgSHO
Move electron pairs, not atoms
Wrong
Write curved arrows to show the bonding changes in this reaction.
Procedure: locate the reactant atoms in the product, and determine the bonds made and broken during the reaction. Then draw curved arrows to show the movement of electron pairs.
Numbering often helps

Labels on curved arrows
a Start here because OH − OH − OH^(-)\mathrm{OH}^{-}OH−has a charge and we need to make the O − C O − C O-C\mathrm{O}-\mathrm{C}O−C bond.
b To preserve the octet on C3 and to make the C1C2 p bond.
c To preserve the octet on C 1 and put the charge on O .
Robert Robinson
Curved arrows, 1924